Question: Rationalize the denominator of $\displaystyle \frac{1}{\sqrt{2} + \sqrt{3} + \sqrt{7}}$, and write your answer in the form \[
\frac{A\sqrt{2} + B\sqrt{3} + C\sqrt{7} + D\sqrt{E}}{F},
\]where everything is in simplest radical form and the fraction is in lowest terms, and $F$ is positive. What is $A + B + C + D + E + F$?
Answer: Since 2, 3, and 7 are all primes, the denominator is in simplest radical form and we can't simplify it further. We attack this problem by getting rid of the square roots one step at a time. First we group the first two terms, and multiply numerator and denominator by the conjugate: \begin{align*}
\frac{1}{(\sqrt{2} + \sqrt{3}) + \sqrt{7}} & = \frac{1}{(\sqrt{2} + \sqrt{3}) + \sqrt{7}} \cdot \frac{(\sqrt{2} + \sqrt{3}) - \sqrt{7}}{(\sqrt{2} + \sqrt{3}) - \sqrt{7}} \\
& = \frac{(\sqrt{2} + \sqrt{3}) - \sqrt{7}}{(\sqrt{2} + \sqrt{3})^2 - (\sqrt{7})^2} \\
& = \frac{(\sqrt{2} + \sqrt{3}) - \sqrt{7}}{2 + 2\sqrt{6} + 3 - 7} \\
& = \frac{\sqrt{2} + \sqrt{3} - \sqrt{7}}{-2 + 2\sqrt{6}}
\end{align*}Now this is in a form we know how to deal with, and we can just multiply by the conjugate as usual: \begin{align*}
\frac{\sqrt{2} + \sqrt{3} - \sqrt{7}}{-2 + 2\sqrt{6}} & = \frac{\sqrt{2} + \sqrt{3} - \sqrt{7}}{-2 + 2\sqrt{6}} \cdot \frac{-2 - 2\sqrt{6}}{-2 - 2\sqrt{6}} \\
& = \frac{-2\sqrt{2} - 2\sqrt{3} + 2\sqrt{7} - 2\sqrt{12} - 2\sqrt{18} + 2\sqrt{42}}{-20} \\
& = \frac{4\sqrt{2} + 3\sqrt{3} - \sqrt{7} - \sqrt{42}}{10}.
\end{align*}This gives $A + B + C + D + E + F = 4 + 3 - 1 - 1 + 42 + 10 = \boxed{57}$.